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Base leg is the diameter of circle,so the center of the circle is the mid of the base.The circle also interecpts the hypo.Draw a line jiong this intersecting point on hypo and to the base mid point, this forms the second triangle which is similar to the original right angle triangle.notice the angles are same as the first one.
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[quote="GoalStanford"]Base leg is the diameter of circle,so the center of the circle is the mid of the base.The circle also interecpts the hypo.Draw a line jiong this intersecting point on hypo and to the base mid point, this forms the second triangle which is similar to the original right angle triangle.notice the angles are same as the first one.[/quote]


I'm not usre if these are similar triangles. becoz
if they are similar the sides should be in the same ratio.
so if the 2 triangles are similar the side that is the diameter to the circle would yield a ratio of 1:2 but the ratio on the hypotenuse would be 4: (4 + 1) which is not equal. I think to prove triangles to be similar the side have to be in the same ratio . Correct me if I'm wrong....
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My apologies,I think I gave wrong triangle there.I was talking about the triangle formed by the two vertices of base leg and the intersecting point on hypo.I will try explain here.

Original triangle ABC , angle B=90 degrees, BC is the base leg.

triangle BDC where D is the intersecting point on hypo.

Now ,
1.angle (ACB)=angle (DCF) no doubts here.you are actually referring to same angle.
2.angle (ABC)=90 degrees. Also, angle (BDC) will be 90 degrees from the cirle theory( angle in semicircle is right angle).

That is how angles are same in both the triangles.



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